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Let $|x - 1| = t$
$t^2 - 5t + 6 = 0$
$t = 2$ & $t = 3$
$|x - 1| = 2$ & $|x - 1| = 3$
$x - 1 = \pm 2,; x - 1 = \pm 3$
$x = 3, -1, 4, -2$
Sum of roots $= 3 + (-1) + 4 + (-2) = 4$
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